Question: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(3t^{3},12t^{2}\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=3$ and $t=8$ ? Round to the nearest tenth.
Answer: To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=3$ and $t=8$ : $\Delta x=\int_{3}^{8} 3t^{3}\,dt=\dfrac{12{,}045}{4}$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=3$ and $t=8$ : $\Delta y=\int_{3}^{8} 12t^{2}\,dt=1940$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(\dfrac{12045}{4}\right)^2+1940^2} \\\\ &\approx 3582.1 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=3$ and $t=8$ is $3582.1$ units.